# 2.3: Density

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- 85140

Density is a physical property that is defined as a substance’s mass divided by its volume:

\[ \begin{align} \text{density} &= \dfrac{\text{mass}}{\text{volume}} \label{eq1} \\[4pt] d &= \dfrac{m}{V} \label{eq2} \end{align}\]

Density is usually a measured property of a substance, so its numerical value affects the significant figures in a calculation. Notice that density is defined in terms of two dissimilar units, mass and volume. That means that density overall has *derived units*, just like velocity. Common units for density include g/mL, g/cm^{3}, g/L, kg/L, and even kg/m^{3}. Densities for some common substances are listed in Table \(\PageIndex{1}\). **Memorize the density of water with its appropriate units**.

Substance |
Density (g/mL or g/cm^{3}) |
---|---|

water |
1.0 |

gold | 19.3 |

mercury | 13.6 |

air | 0.0012 |

cork | 0.22–0.26 |

aluminum | 2.7 |

iron | 7.87 |

Because of how it is defined, density can act as a conversion factor for switching between units of mass and volume. For example, suppose you have a sample of aluminum that has a volume of 7.88 cm^{3}. How can you determine what mass of aluminum you have without measuring it? You can use the volume to calculate it. If you multiply the given volume by the known density (from Table \(\PageIndex{1}\)), the volume units will cancel and leave you with mass units, telling you the mass of the sample:

Start with Equation \ref{eq1} \[\text{density} = \dfrac{m}{V} \nonumber\]

and insert the relavant numbers

\[\dfrac{2.7g}{cm^3} = \dfrac{m}{7.88 \, cm^3} \nonumber\]

Cross multiplying both sides (right numerator x left denominator = left numerator x right denominator), we get the following expression with answer and appropriate unit.

\[7.88\cancel{cm^3}\times \dfrac{2.7\,g}{\cancel{cm^3}}= 21 g \text{ of aluminum} \nonumber\]

Example \(\PageIndex{1}\): Mercury

What is the mass of 44.6 mL of mercury?

**Solution**

Use the density value for mercury from Table \(\PageIndex{1}\) and the definition of density (Equation \ref{eq1})

\[density= \dfrac{mass}{volume}\Rightarrow d= \dfrac{m}{V} \nonumber\]

\[\dfrac{13.6g}{mL}= \dfrac{m}{44.6 \, mL} \nonumber\]

Remember to cross multiply here in order to isolate variable. Then, report answer with correct units.

\[44.6\cancel{mL}\times \dfrac{13.6\,g}{\cancel{mL}}= 607\,g \nonumber\]

The mass of the mercury is 607 g.

Exercise \(\PageIndex{1}\)

What is the mass of 25.0 cm^{3} of iron?

**Answer**-
Use the density value for iron from Table \(\PageIndex{1}\)

\[density= \dfrac{mass}{volume}\Rightarrow d= \dfrac{m}{V} \nonumber\]

\[\dfrac{7.87g}{cm^3}= \dfrac{m}{25.0\, cm^3} \nonumber\]

Cross multiplying both sides (right numerator x left denominator = left numerator x right denominator), we get the following expression with answer and appropriate unit.

\[25.0\cancel{cm^3}\times \dfrac{7.87\,g}{\cancel{cm^3}}= 197 g \text{ of iron} \nonumber\]

**Another way of looking at density (some students choose to perform calculations using this method)**

Density can also be used as a conversion factor to convert mass to volume—but care must be taken. We have already demonstrated that the number that goes with density normally goes in the numerator when density is written as a fraction. Take the density of gold, for example:

\[d=19.3\,g/mL =\dfrac{19.3\,g}{mL} \nonumber\]

Although this was not previously pointed out, it can be assumed that there is a 1 in the denominator:

\[d=19.3\,g/mL =\dfrac{19.3\,g}{mL} \nonumber\]

That is, the density value tells us that we have 19.3 grams for every 1 milliliter of volume, and the 1 is an exact number. When we want to use density to convert from mass to volume, the numerator and denominator of density need to be switched—that is, we must take the *reciprocal* of the density. In so doing, we move not only the units but also the numbers:

\[\dfrac{19.3\,\cancel{g}}{mL}= \dfrac{45\,\cancel{g}}{V} \nonumber\]

Cross multiplying denominators with numerators, we obtain the following algebraic equation.

\[19.3 V = 45.9\, mL \nonumber\]

then you will need to isolate the variable (volume)

\[ V = \dfrac{ 45.9\,mL}{19.3} \nonumber\]

After multiplication, the answer would be

\[ V = 2.38 \,mL \nonumber\]

Example \(\PageIndex{2}\): Wine Cork

A cork stopper from a bottle of wine has a mass of 3.78 g. If the density of cork is 0.22 g/mL, what is the volume of the cork?* Regardless of the method that is used, you should still be able to obtain the same (and correct) answer. *

**Solution**

To use density as a conversion factor, we need to take the reciprocal so that the mass unit of density is in the denominator. Taking the reciprocal, we find

\[\dfrac{0.22\,\cancel{g}}{mL}= \dfrac{3.78\,\cancel{g}}{V} \nonumber\]

Cross multiplying denominators with numerators, we obtain the following algebraic equation.

\[0.22 V = 3.78\,mL \nonumber\]

then you will need to isolate the variable (volume)

\[ V = \dfrac{3.78\,mL}{0.22} \nonumber\]

so, the volume of the cork is 17.2 mL.

Exercise \(\PageIndex{2}\)

What is the volume of 3.78 g of gold?

**Answer**-
Before attempting this question, be sure to obtain the density of gold in the table above. If you were to need this value on a quiz or a test, then it would be provided. Once you have this value, plug it into the density equation. Next, you will need to isolate the volume variable (basic algebra). The final answer should be 0.196 cm

^{3}.

Care must be used with density as a conversion factor. Make sure the mass units are the same or the volume units are the same, before using density to convert to a different unit. Often, the unit of the given quantity must be first converted to the appropriate unit before applying density as a conversion factor.

## Using Density in Environmental Applications

Along with solubility, density can help determine how a contaminant could affect an aquatic system. For example, imagine mercury has been spilled in Furman Lake. Looking at this element's density value and comparing it to liquid water, one could determine the location of the insoluble (you would be given solubility information) mercury layer. The more dense mercury layer would reside on the bottom of Furman Lake. If one were to take a cross-section of the lake, you could see that a heterogeneous mixture would result.

In contrast, spilling ethanol (density = 0.789g/mL) would result in the formation of a homogeneous mixture. Ethanol (grain alcohol) is soluble in water. This would make it miscible (mixable to form a solution) in water and one would not be able to denote separate layers. According to the density, an alcohol layer would remain on top, but would ultimately dissolve.

Applications

What difficulties would arise from the separation and removal of contaminants?

- Hg in Furman Lake
- Ethanol in Furman Lake
- oil (less dense, insoluble) in Furman Lake

Watch this video and record your observations.

- What component was different in the two types of beverages (mass or volume)?
- How does the above-mentioned difference affect the density equation?
- Which beverage is denser than water?

## Need More Practice?

- Turn to Section 2.E of this OER and work problems #2 and #9.